On the 𝑋 -coordinates of Pell equations which are rep-digits, II
Florian Luca
a, Sossa Victorin Togan
b, Alain Togbé
caSchool of Mathematics, University of the Witwatersrand, South Africa and Department of Mathematics, Faculty of Sciences, University of Ostrava, Czech Republic
florian.luca@wits.ac.za
bInstitut de Mathématiques et de Sciences Physiques, Porto-Novo, Bénin tofils74@yahoo.fr
cDepartment of Mathematics, Statistics and Computer Science Purdue University Northwest,Westville,USA
atogbe@pnw.edu Submitted: June 27, 2018 Accepted: December 12, 2018 Published online: February 11, 2019
Abstract
For a positive integer 𝑑 which is not a square, we show that there is at most one value of the positive integer𝑋 participating in the Pell equation 𝑋2−𝑑𝑌2=±4which is a rep-digit, that is all its base10digits are equal, except for𝑑= 2,5,13.
Keywords:Pell equation, Rep-digit, Linear forms in complex logarithms.
MSC:11A25 11B39, 11J86
1. Introduction
Let𝑑be a positive integer which is not a perfect square. It is well-known that the Pell equation
𝑋2−𝑑𝑌2=±4 (1.1)
has infinitely many positive integer solutions(𝑋, 𝑌). Furthermore, putting(𝑋1, 𝑌1) for the smallest such solution (solution with minimal value for𝑋), all the positive doi: 10.33039/ami.2018.12.002
http://ami.uni-eszterhazy.hu
131
integer solutions are of the form(𝑋𝑛, 𝑌𝑛)for some positive integer𝑛where 𝑋𝑛+√
𝑑𝑌𝑛
2 =
(︃𝑋1+√ 𝑑𝑌1
2
)︃𝑛
.
There are many papers in the literature which solve Diophantine equations involv- ing members of the sequences{𝑋𝑛}𝑛≥1or{𝑌𝑛}𝑛≥1being squares, or perfect powers of larger exponents of some other integers, etc. (see, for example, [4, 5]).
Let𝑔≥2 be an integer. A natural number𝑁 is called a base𝑔 rep-digitif all of its base𝑔-digits are equal; that is, if
𝑁 =𝑎
(︂𝑔𝑚−1 𝑔−1
)︂
, for some 𝑚≥1and𝑎∈ {1,2, . . . , 𝑔−1}.
When𝑔= 10, we omit the base and simply say that𝑁 is a rep-digit. Diophantine equations involving rep-digits were also considered in several papers which found all rep-digits which are perfect powers, or Fibonacci numbers, or generalized Fibonacci numbers, and so on (see [1–3, 7, 9, 11–15, 17] for a sample of such results). In this paper, we study when can 𝑋𝑛 be a rep-digit. This reduces to the Diophantine equation
𝑋𝑛 =𝑎
(︂10𝑚−1 9
)︂
, 𝑚≥1and𝑎∈ {1, . . . ,9}. (1.2) Of course, for every positive integer𝑋, there is a unique square-free integer𝑑≥2 such that
𝑋2−𝑑𝑌2=−4.
Namely𝑑is the product of all prime factors of𝑋2+4which appear at odd exponents in its factorization. In particular, taking𝑋 =𝑎(10𝑚−1)/9, we get that any rep- digit is the𝑋-coordinate of the Pell equation (1.1) corresponding to some specific square-free integer𝑑. If𝑋 >2, we can instead look at𝑋2−4and write it as𝑑𝑌2 for some positive integers𝑑and𝑌 with𝑑squarefree, and then
𝑋2−𝑑𝑌2= 4.
In particular, we can take𝑋 =𝑎(10𝑚−1)/9with𝑎∈ {1, . . . ,9}and𝑚≥1, where we ask in addition that𝑎≥3when𝑚= 1. Here, we study the square-free integers 𝑑 such that the sequence{𝑋𝑛}𝑛≥1 contains at least two rep-digits. Our result is the following.
Theorem 1.1. Let 𝑑≥2 be square-free. The Diophantine equation 𝑋𝑛=𝑎
(︂10𝑚−1 9
)︂
, 𝑚≥1 and𝑎∈ {1, . . . ,9} (1.3) has at most one positive integer solution 𝑛 except when 𝑑= 2,5,13 for which we have
22−2·22=−4, 62−2·42= 4,
12−5·12=−4, 32−5·12= 4, 42−5·22=−4 72−5·32= 4, 112−5·52=−4, and
32−13·12=−4, 112−13·32= 4.
2. Linear forms in logarithms
We need some results from the theory of lower bounds for nonzero linear forms in logarithms of algebraic numbers. We start by recalling Theorem 9.4 of [4], which is a modified version of a result of Matveev [16]. LetLbe an algebraic number field of degree 𝑑L. Let 𝜂1, 𝜂2, . . . , 𝜂𝑙 ∈L not0 or 1 and 𝑑1, . . . , 𝑑𝑙 be nonzero integers.
We put
𝐷= max{|𝑑1|, . . . ,|𝑑𝑙|,3}, and
Γ =
∏︁𝑙 𝑖=1
𝜂𝑑𝑖𝑖−1.
Let𝐴1, . . . , 𝐴𝑙 be positive integers such that
𝐴𝑗 ≥ℎ′(𝜂𝑗) := max{𝑑Lℎ(𝜂𝑗),|log𝜂𝑗|,0.16}, for 𝑗= 1, . . . 𝑙, where for an algebraic number𝜂 of minimal polynomial
𝑓(𝑋) =𝑎0(𝑋−𝜂(1))· · ·(𝑋−𝜂(𝑘))∈Z[𝑋]
over the integers with positive𝑎0, we write ℎ(𝜂)for its Weil height given by
ℎ(𝜂) = 1 𝑘
⎛
⎝log𝑎0+
∑︁𝑘 𝑗=1
max{0,log|𝜂(𝑗)|}
⎞
⎠.
The following consequence of Matveev’s theorem is Theorem 9.4 in [4].
Theorem 2.1. If Γ̸= 0andL⊆R, then
log|Γ|>−1.4·30𝑙+3𝑙4.5𝑑2L(1 + log𝑑L)(1 + log𝐷)𝐴1𝐴2· · ·𝐴𝑙.
When𝑙 = 2 and 𝜂1, 𝜂2 are positive and multiplicatively independent, we can do better. Namely, let in this case𝐵1, 𝐵2 be real numbers larger than1such that
log𝐵𝑖≥max {︂
ℎ(𝜂𝑖),|log𝜂𝑖| 𝑑L , 1
𝑑L }︂
𝑖= 1,2, and put
𝑏′:= |𝑑1| 𝑑Llog𝐵2
+ |𝑑2| 𝑑Llog𝐵1
. Furthermore, let
Λ =𝑑1log𝜂1+𝑑2log𝜂2.
Note thatΛ̸= 0 when𝜂1 and𝜂2 are multiplicatively independent.
Theorem 2.2. With the above notations, assuming that L is real, 𝜂1, 𝜂2 are positive and multiplicatively independent, then
log|Λ|>−24.34𝑑4L (︂
max {︂
log𝑏′+ 0.14,21 𝑑L,1
2 }︂)︂2
log𝐵1log𝐵2.
Note that𝑒Λ−1 = Γ, soΓis close to zero if and only ifΛis close to zero, which explains the relation between Theorems 2.1 and 2.2.
3. The Baker-Davenport lemma
Here, we recall the Baker-Davenport reduction method (see [8, Lemma 5a]), which turns out to be useful in order to reduce the bounds arising from applying Theorems 2.1 and 2.2.
Lemma 3.1. Let 𝜅 ̸= 0 and 𝜇 be real numbers. Assume that 𝑀 is a positive integer. Let 𝑃/𝑄 be the convergent of the continued fraction expansion of𝜅 such that 𝑄 >6𝑀 and put
𝜉=‖𝜇𝑄‖ −𝑀· ‖𝜅𝑄‖,
where‖ · ‖denotes the distance from the nearest integer. If 𝜉 >0, then there is no solution to the inequality
0<|𝑚𝜅−𝑛+𝜇|< 𝐴𝐵−𝑘 in positive integers𝑚,𝑛and 𝑘with
log (𝐴𝑄/𝜉)
log𝐵 ≤𝑘 and 𝑚≤𝑀.
4. Bounding the variables
We assume that(𝑋1, 𝑌1)is the minimal solution of the Pell equation (1.1). Set 𝑋12−𝑑𝑌12=:±4
and
𝑥𝑛 =𝑋𝑛
2 , 𝑦𝑛 =𝑌𝑛
2 for all 𝑛≥1.
We have
𝑥2𝑛−𝑑𝑦𝑛2=:𝜀𝑛, 𝜀𝑛∈ {±1}. Put
𝛿:=𝑥1+
√︁
𝑥21−𝜀1=𝑥1+√
𝑑𝑦1, 𝜂:=𝑥1−√
𝑑𝑦1=𝜀1𝛿−1, with 𝛿≥(1+√ 5)/2.
Then, we get
𝑥𝑛 =1
2(𝛿𝑛+𝜂𝑛), or, equivalently,
𝑋𝑛=𝛿𝑛+𝜂𝑛.
We start with some general considerations concerning equation (1.2). From equa- tion (1.2), we have
𝑋𝑛=𝑎
(︂10𝑚−1 9
)︂
> 𝑎(1 + 10 +· · ·+ 10𝑚−1)>10𝑚−1. We get
10𝑚−1≤𝑋𝑛<10𝑚. (4.1)
Furthermore,
2𝛿𝑛> 𝛿𝑛+𝜂𝑛=𝑋𝑛≥𝛿𝑛−𝛿−𝑛 ≥𝛿𝑛 2 , where the last inequality follows because𝑛≥1 and𝛿≥(1 +√
5)/2>√ 2. So, 𝛿𝑛
2 ≤𝑋𝑛 <2𝛿𝑛 holds for all 𝑛≥1. (4.2) Using now the equations (4.1) and (4.2), we have
10𝑚−1≤𝑋𝑛<2𝛿𝑛 and 𝛿𝑛
2 ≤𝑋𝑛 ≤10𝑚. Hence, we obtain
𝑛𝑐1log𝛿−𝑐2≤𝑚≤𝑛𝑐1log𝛿+𝑐2+ 1, 𝑐1:= 1/log 10, 𝑐2:=𝑐1log 2. (4.3) From the left-hand side inequality of (4.3), we also deduce that
𝑛log𝛿 < 𝑚log 10 + log 2. (4.4)
Since𝛿≥(1 +√
5)/2, we get that 𝑛≤𝑚 log 10
log((1 +√
5)/2) + log 2 log((1 +√
5)/2) <4.8𝑚+ 2.
If 𝑚≥2, the last inequality above implies that𝑛 <6𝑚. If𝑚= 1, then𝑋𝑛 ≤9, so 𝛿𝑛 ≤18 by (4.2). Since𝛿 ≥(1 +√
5)/2, we get that 𝑛≤6, so the inequality 𝑛≤6𝑚holds also when𝑚= 1. We record this as
𝑛≤6𝑚. (4.5)
Next, using (1.3), we get
𝛿𝑛+𝜂𝑛=𝑎
(︂10𝑚−1 9
)︂
.
Put 𝑏:=𝑎/9. We have
𝛿𝑛𝑏−110−𝑚−1 =−𝑏−110−𝑚𝜂𝑛−10−𝑚. Thus,
⃒⃒𝛿𝑛𝑏−110−𝑚−1⃒⃒≤ 1
𝑏10𝑚𝛿𝑛 + 1 10𝑚 = 1
10𝑚 (︂
1 + 9 𝑎𝛿𝑛
)︂
< 6 10𝑚, using that𝑎≥1, 𝑛≥1and𝛿≥(1 +√
5)/2. Thus,
⃒⃒𝛿𝑛𝑏−110−𝑚−1⃒⃒< 6
10𝑚. (4.6)
We now assume that 𝑚≥2 and search for an upper bound on it. Since𝑚≥2, it follows that the right-hand side in (4.6) above is<1/2. Put
Λ :=𝑛log𝛿−log𝑏−𝑚log 10.
Since|𝑒Λ−1|<1/2, it follows that
|Λ|<2|𝑒Λ−1|< 12 10𝑚. Let us return to (4.6) and put
Γ :=𝑒Λ−1 =𝛿𝑛𝑏−110−𝑚−1.
Note that Γ is nonzero. Indeed, if it were zero, then𝛿𝑛 =𝑏10𝑚. Hence, 𝛿𝑛 ∈Q.
Since𝛿 is an algebraic integer and𝑛≥1, it follows that𝛿𝑛 ∈Z. Since𝛿is a unit, we get that 𝛿𝑛 = 1, so 𝑛= 0, which is a contradiction. Thus, Γ ̸= 0. We apply Matveev’s theorem. If𝑎̸= 9(so, 𝑏̸= 1), we then take
𝑙= 3, 𝜂1=𝛿, 𝜂2=𝑏, 𝜂3= 10, 𝑑1=𝑛, 𝑑2=−1, 𝑑3=−𝑚, 𝐷= max{𝑛, 𝑚}. Clearly,L=Q[√
𝑑]contains all the numbers𝜂1, 𝜂2, 𝜂3 and has degree𝑑L= 2. We have
ℎ(𝜂1) = (1/2) log𝛿, ℎ(𝜂2)≤log 9 and ℎ(𝜂3) = log 10.
Thus, we can take
𝐴1= log𝛿, 𝐴2= 2 log 9 and 𝐴3= 2 log 10.
Now, Theorem 2.1 tells us that
log|Γ|>−1.4×306×34.5×22(1 + log 2)(1 + log𝐷)(log𝛿)(2 log 9)(2 log 10).
Comparing the above inequality with (4.6), we get
𝑚log 10−log 6<1.4×306×34.5×24(1 + log 2)(1 + log𝐷)(log𝛿)(log 9)(log 10).
Thus,
𝑚 <1.4×306×34.5×24×(log 9)(1 + log 2)×(log𝛿)·(1 + log𝐷) or
𝑚 <8.6·1012(1 + log𝐷) log𝛿.
Since𝐷≤6𝑚(see (4.5)), we get
𝑚 <8.6·1012(1 + log(6𝑚)) log𝛿. (4.7) This was when𝑏̸= 1. In case𝑏= 1, we take𝑙 = 2and apply the same inequality (except that now𝜂2:= 1is no longer present) getting a better result. Finally, this was under the assumption that𝑚≥2but if𝑚= 1then inequality (4.7) also holds.
Let us record what we have proved so far.
Lemma 4.1. Denoting by𝛿:=𝑥1+√
𝑑𝑦1, all positive integer solutions(𝑚, 𝑛)of equation (1.2)satisfy
𝑚 <8.6·1012(1 + log(6𝑚)) log𝛿.
All this is for the equation𝑋𝑛=𝑎(10𝑚−1)/9. Now we assume that 𝑋𝑛1=𝑎1
(︂10𝑚1−1 9
)︂
and 𝑋𝑛2 =𝑎2
(︂10𝑚2−1 9
)︂
. where𝑎1, 𝑎2∈ {1, . . . ,9}.
To fix ideas, we assume that𝑛1< 𝑛2, so𝑚1≤𝑚2. We put as before𝑏𝑖:=𝑎𝑖/9 for𝑖= 1,2. From the above analysis, assuming that𝑚1≥2, we have that
|𝑛𝑖log𝛿−log𝑏𝑖−𝑚𝑖log 10|< 12
10𝑚𝑖 holds for 𝑖∈ {1,2}. (4.8) The argument proceeds in two steps according to whether𝑏1𝑏2<1or𝑏1𝑏2= 1.
Suppose now that𝑏1𝑏2<1.
We multiply the equation (4.8) for𝑖= 1with𝑛2and the one for𝑖= 2with𝑛1, subtract them and apply the absolute value inequality to get
|𝑛2log𝑏1−𝑛1log𝑏2+ (𝑛2𝑚1−𝑛1𝑚2) log 10| (4.9)
=|𝑛1(𝑛2log𝛿−log𝑏2−𝑚2log 10)−𝑛2(𝑛1log𝛿−log𝑏1−𝑚1log 10)|
≤𝑛1|𝑛2log𝛿−log𝑏1−𝑚2log 10|+𝑛2|𝑛1log𝛿−log𝑏1−𝑚1log 10|
≤ 12𝑛1
10𝑚2 +12𝑛2
10𝑚1 ≤ 24𝑛2
10𝑚1.
If the right-hand side above is at least1/2, we then get 10𝑚1 ≤48𝑛2<300𝑚2, giving
𝑚1< 𝑐1log(300𝑚2). (4.10)
Assume now that the right-hand side in (4.9) is smaller than1/2. Putting, Λ0:=𝑛2log𝑏1−𝑛1log𝑏2+ (𝑛2𝑚1−𝑛1𝑚2) log 10,
we get|Λ0|<1/2. Putting
Γ0:=𝑏𝑛12𝑏−2𝑛110𝑛2𝑚1−𝑛1𝑚2−1, we get that
|Γ0|=|𝑒Λ0−1|<2|Λ0|< 48𝑛2
10𝑚1, (4.11)
where the middle inequality above follows from the fact that|Λ0|<1/2. We apply Matveev’s theorem to estimate a lower bound onΓ0. But first, let us see that it is nonzero. AssumingΓ0= 0, we get
𝑏𝑛12𝑏−2𝑛1= 10𝑛2𝑚1−𝑛1𝑚2. (4.12) Assume first that 𝑛2𝑚1−𝑛1𝑚2 = 0. Then 𝑏𝑛12 = 𝑏𝑛21. Thus, 𝑏1 and 𝑏2 are multiplicatively independent and they belong to the set
{︂1 9,2
9,1 3,4
9,5 9,2
3,7 9,8
9,1 }︂
.
They are not both1and𝑛1and𝑛2are both positive. So, the only possibilities are that 𝑏1=𝑏2, or
{𝑏1, 𝑏2}= {︂1
9,1 3
}︂
, {︂2
3,4 9
}︂
. (4.13)
If 𝑏1 = 𝑏2, then 𝑏𝑛11 = 𝑏𝑛22 implies 𝑛1 = 𝑛2, which together with 𝑛2𝑚1 = 𝑛1𝑚2
leads to𝑚1=𝑚2. Thus,(𝑛1, 𝑚1) = (𝑛2, 𝑚2)and 𝑎1=𝑎2 (because𝑏1=𝑏2), and this is not convenient for us. If{𝑏1, 𝑏2}is one of the two sets from (4.13), then one of𝑏1, 𝑏2 is the square of the other one. Thus, since𝑏𝑛11 =𝑏𝑛22 and𝑛2> 𝑛1, we get 𝑛2 = 2𝑛1. Since also 𝑛2𝑚1 =𝑛1𝑚2, we have 𝑚2= 2𝑚1. Hence, also𝑏2 =𝑏21 and 𝑏1∈ {1/3,2/3}. So, we get the pair of equations
𝑋𝑛1 =𝑏110𝑚1−𝑏1 and 𝑋2𝑛1 =𝑏21102𝑚1−𝑏21. Since in fact
𝑋2𝑛=𝛿2𝑛+𝜂2𝑛= (𝛿𝑛+𝜂𝑛)2−2(𝛿𝜂)𝑛=𝑋𝑛2±2,
we get that
𝑏21102𝑚1−𝑏21=𝑋2𝑛1 =𝑋𝑛21±2 = (𝑏110𝑚1−𝑏1)2±2 =𝑏21102𝑚1−2𝑏2110𝑚1+𝑏21±2, which leads to
2𝑏2110𝑚1 = 2𝑏21±2, so
10𝑚1= 1±𝑏−12.
The last equation above is impossible for𝑚1≥2. For𝑚1= 1we get10 = 1±𝑏−12, which gives 𝑏1= 1/3. Hence,
𝑋𝑛1= 10−1
3 = 3, and 𝑋2𝑛1 = 102−1 9 = 11.
Since 𝑋2𝑛1 =𝑋𝑛21±2, it follows that the sign is +, so 𝑋𝑛21−𝑑𝑌𝑛21 =−4, giving 𝑑𝑌𝑛21 = 13, so 𝑑 = 13, 𝑌1 = 1, 𝑛1 = 1. These solutions are among the ones mentioned in the statement of the main theorem.
This deals with the case when𝑛2𝑚1−𝑛1𝑚2 = 0. Assume next that 𝑛2𝑚1− 𝑛1𝑚2̸= 0. Then in the right-hand side of (4.12), both primes2and5are involved at a nonzero exponent. Thus, they should be also involved with nonzero exponents in the left-hand side of (4.12). Thus, one of 𝑏1, 𝑏2 is 5/9 and the other is in {2/9, 4/9, 2/3, 8/9}. A minute of reflection shows that in all cases the exponents of2 and5 in the left-hand side of (4.12) have opposite signs, whereas in the right they have the same sign, and this is impossible.
Thus,Γ0̸= 0. Hence, we are entitled to apply Matveev’s theorem in order to find a lower bound on Γ0. In case𝑏1̸= 1and𝑏2̸= 1, we take
𝑙= 3, 𝜂1=𝑏1, 𝜂2=𝑏2, 𝜂3= 10, 𝑑1=𝑛2, 𝑑2=−𝑛1, 𝑑3=𝑛2𝑚1−𝑛1𝑚2. Clearly,L=Qcontains all the numbers𝜂1, 𝜂2, 𝜂3and has degree𝑑L= 1. Further, 𝐷= max{|𝑑1|,|𝑑2|,|𝑑3|} ≤𝑛2𝑚2≤6𝑚22. We have
ℎ(𝜂1)≤log 9, ℎ(𝜂2)≤log 9 and ℎ(𝜂3) = log 10.
Thus, we can take
𝐴1= log 9, 𝐴2= log 9, 𝐴3= log 10.
Now, Theorem 2.1 tells us that
log|Γ0|>−1.4×306×34.5(1 + log𝐷)(log 9)2(log 10).
Combining this with estimate (4.11) and using the fact that 48𝑛2 <300𝑚2 (see inequality (4.5)) we get
𝑚1log 10≤log 300 + log𝑚2+ 1.6×1012(1 + log(6𝑚22)),
giving
𝑚1<7×1011(1 + log(6𝑚22)). (4.14) The right-hand side of inequality (4.14) is larger than the right-hand side of in- equality (4.10). So, regardless whether 24𝑛2/10𝑚1 is at least 1/2 or smaller than 1/2, estimate (4.14) holds. From equation (4.4), we get
log𝛿 <(𝑚1+ 1) log 10<1.7×1012(1 + log(6𝑚22)), which together with Lemma 4.1 gives
𝑚2<(︀
8.6×1012(1 + log(6𝑚2)))︀ (︀
1.7×1012(1 + log(6𝑚22)))︀
, so
𝑚2<1.5×1025(1 + log(6𝑚2))(1 + log(6𝑚22)).
This gives𝑚2<1.5×1029. This was if both𝑏1 and𝑏2are different than1. If one of them is1, we simply apply Matveev’s theorem with𝑙= 2getting an even better bound for𝑚2.
Suppose now that𝑏1=𝑏2= 1.
We return to (4.11) getting that8/9 ≤24𝑛2/10𝑚1, which leads to (4.10), un- less 𝑛1𝑚2 =𝑛2𝑚1. In this last case, we get that 𝑛2/𝑚2 =𝑛1/𝑚1. Thus, writ- ing 𝑛1/𝑚1 = 𝑟/𝑠 in reduced terms, we get that (𝑛1, 𝑚1) = (ℓ1𝑟, ℓ1𝑠) and that (𝑛2, 𝑚2) = (ℓ2𝑟, ℓ2𝑠)for some positive integersℓ1< ℓ2. Hence, we have
𝑋𝑟ℓ1 = 10𝑠ℓ1−1, 𝑋𝑟ℓ2 = 10𝑠ℓ2−1.
The greatest common divisor of the right hand sides above is 10𝑠−1 ≥9. The greatest common divisor of the left-hand sides above is𝑋𝑟 ifℓ1ℓ2 is odd and 1or 2 otherwise. Thus,ℓ1ℓ2 must be odd and
𝑋𝑟= 10𝑠−1.
Consequently,
𝛿𝑟−10𝑠=−𝜂𝑟−1 and 𝛿ℓ2𝑟−10ℓ2𝑠=−𝜂ℓ2𝑟−1.
From the two equations above we get
𝛿(ℓ2−1)𝑟+𝛿(ℓ2−2)10𝑠+· · ·+ 10(ℓ2−1)𝑠= −𝜂ℓ2𝑟−1
−𝜂𝑟−1 .
The last relation above is impossible since its left-hand side is >10and its right hand side is
≤ 2
1−1+2√5 <10, a contradiction.
In conclusion, (4.10) holds, which is stronger than (4.14), and the above argu- ments imply that𝑚2<1.5×1029. Hence, we have the following result.
Lemma 4.2. The inequality
𝑚2<1.5×1029 holds.
Now one needs to apply LLL to the bound
|Λ0|< 24𝑛2
10𝑚1 <24×6×1.5×1029
10𝑚1 < 1 10𝑚1−32 to get a reasonably small bound on𝑚1.
∙First, we will consider the case𝑏1=𝑏2:=𝑏; i.e.,𝑎1=𝑎2:=𝑎or {𝑏1, 𝑏2} ∈
{︂1 9,1
3 }︂
, {︂2
3,4 9
}︂
. In
Λ0:=𝑛2log𝑏1−𝑛1log𝑏2+ (𝑛2𝑚1−𝑛1𝑚2) log 10, (4.15) we set 𝑋 :=𝑛1−𝑛2 or 𝑋 := 2𝑛2−𝑛1, and𝑌 :=𝑛2𝑚1−𝑛1𝑚2 and divide both sides by𝑌log𝑏 (with𝑏=𝑏1=𝑏2∈ {1/9,2/9,3/9,4/9,5/9,6/9,7/9,8/9}) to get
⃒⃒
⃒⃒log 10 log𝑏 −𝑋
𝑌
⃒⃒
⃒⃒< 1
𝑌(log(1/𝑏))10𝑚1−32. (4.16) We assume that 𝑚1 is so large that the right-hand side in (4.16) is smaller than 1/(2𝑌2). This certainly holds if
10𝑚1−32>2(log(1/𝑏))−1𝑌. (4.17) Since|𝑌|<1.5×1059, it follows that the last inequality (4.17) holds provided that 𝑚1 ≥92 in all cases, which we now assume. In this case,𝑋/𝑌 is a convergent of the continued fraction of𝜂 := log 10/log𝑏 and𝑋 <1.5×1059. Writing
𝑎= 1, 𝜂:= [−2,1,19,1,5,1,6,2,5,15,3, . . . ,7,2,121,1, . . . ,2,569,1,2,27,7, . . .]
𝑎= 2, 𝜂:= [−2,2,7,1,1,2,4,2,99, . . .]1,292,1,6,1,3,3,2,2,5, . . . ,1,1,1,42, . . .]
𝑎= 3, 𝜂:= [−3,1,9,2,2,1,13,1,7,18, . . . ,2,10,3,1,1,1,1,1,6, . . . ,1,284,2, . . .]
𝑎= 4, 𝜂:= [−3,6,4,2,1,1,1,1,45,89,1,6,1,9,1,2,625, . . . ,2,2,1,1716,1,1, . . .]
𝑎= 5, 𝜂:= [−4,12,9,1,1,1,1,1,2,1, . . . ,10,1,1,12,8860,4,13,1,1,5,3,9,1, . . .]
𝑎= 6, 𝜂:= [−6,3,8,1,3,3,22,1,1,44, . . . ,1,1,38,1,5,1,857,1,3,1,3,1,2,1, . . .]
𝑎= 7, 𝜂:= [−10,1,5,6,118,2,8,1,2,1, . . . ,8,23,1,30,2,2,8,1,4,2,1,1,255, . . .]
𝑎= 8, 𝜂:= [−20,2,4,1,1,3,2,7,1,2,1,9,2,6, . . . ,1,2,1332,1,12,1,5,1,1,2, . . .]
for the continued fraction of 𝜂 and 𝑝𝑘/𝑞𝑘 for the 𝑘th convergent, we get that 𝑋/𝑌 =𝑝𝑗/𝑞𝑗 for some𝑗 ≤122 in all cases. Furthermore, putting𝑀 := max{𝑎𝑗: 0 ≤ 𝑗 ≤ 122}, we get 𝑀 = 8860 (for 𝑎= 5). From the known properties of the continued fractions, we then get that
1
8862𝑌2 = 1 (𝑀+ 2)𝑌2 ≤
⃒⃒
⃒⃒𝜂−𝑋 𝑌
⃒⃒
⃒⃒< 1
𝑌(log𝑏)10𝑚1−32, giving
10𝑚1−32<8862(log𝑏)−1𝑌 <8862(log𝑏)−1(1.5×1059), leading to𝑚1≤96.
∙ We now consider the remaining cases. We transform the linear form (4.15) into one of the following forms:
Λ1= (𝑚1𝑛2−𝑚2𝑛1+𝛿1𝑛1+𝛿2𝑛2) log 2 + (𝜆1𝑛1+𝜆2𝑛2) log 3 +(𝑚1𝑛2−𝑚2𝑛1+𝜇1𝑛1+𝜇2𝑛2) log 5,
Λ2:= (𝜆1𝑛1+𝜆2𝑛2) log 3 + (𝜈1𝑛1+𝜈2𝑛2) log 7 + (𝑚1𝑛2−𝑚2𝑛1) log 10, Λ3= (𝑚1𝑛2−𝑚2𝑛1+𝛿1𝑛1+𝛿2𝑛2) log 2 + (𝜆1𝑛1+𝜆2𝑛2) log 3
+(𝑚1𝑛2−𝑚2𝑛1+𝜇1𝑛1+𝜇2𝑛2) log 5 + (𝜈1𝑛1+𝜈2𝑛2) log 7, where|𝛿𝑖| ≤3,|𝜆𝑖| ≤2, |𝜇𝑖| ≤1,|𝜈𝑖| ≤1, for𝑖= 1,2.
Now, we will estimate lower bounds for Λ𝑖, 𝑖 = 1,2,3 via the LLL algorithm (see Proposition 2.3.20 in [6]). One knows that Λ𝑖̸= 0,𝑖= 1,2,3by what is done above. We set𝑋1=𝑋3:= 1060as upper bounds for|𝑚1𝑛2−𝑚2𝑛1+𝛿1𝑛1+𝛿2𝑛2|,
|𝑚1𝑛2−𝑚2𝑛1+𝜇1𝑛1+𝜇2𝑛2|and 𝑋2=𝑋4:= 1031 as upper bounds for|𝜆1𝑛1+ 𝜆2𝑛2|, |𝜈1𝑛1+𝜈2𝑛2|. We take 𝐶 := (3𝑋1)3 for Λ1, Λ2 and 𝐶 := (4𝑋1)4 forΛ3. Moreover, we consider the latticeΩspanned by
𝑣1:= (1,0,⌊𝐶log 2⌋), 𝑣2:= (0,1,⌊𝐶log 3⌋), 𝑣3:= (0,0,⌊𝐶log 5⌋), forΛ1
𝑣1:= (1,0,⌊𝐶log 3⌋), 𝑣2:= (0,1,⌊𝐶log 7⌋), 𝑣3:= (0,0,⌊𝐶log 10⌋), forΛ2
𝑣1:= (1,0,0,⌊𝐶log 2⌋), 𝑣2:= (0,1,0,⌊𝐶log 3⌋), 𝑣3:= (0,0,1,⌊𝐶log 5⌋), 𝑣4:= (0,0,0,⌊𝐶log 7⌋),
forΛ3. Then, we compute𝑄, 𝑇, 𝑐1, 𝑚according to Proposition 2.3.20 in [6] and we obtain:
5.5·10−122<|Λ1|< 1
10𝑚1−32 ⇒ 𝑚1≤153;
3.2·10−122<|Λ2|< 1
10𝑚1−32 ⇒ 𝑚1≤153;
8.1·10−183<|Λ3|< 1
10𝑚1−32 ⇒ 𝑚1≤214.
Hence, we have the following numerical result.
Lemma 4.3. The estimate 𝑚1≤214 holds.
For𝑎1∈ {1,2, . . . ,9},1≤𝑛1≤1284,1≤𝑚1≤214, we solve the equations 𝑥𝑛1=𝑃𝑛1(𝑥1) =𝑎1
(︂10𝑚1−1 9
)︂
to see for which values of the triple (𝑛1, 𝑚1) it has a solution 𝑥1 = 𝑋1/2 with positive integer𝑋1, where
𝑥𝑛=𝑃𝑛(𝑋/2) =
(︃𝑋+√ 𝑋2±4 2
)︃𝑛
+
(︃𝑋−√ 𝑋2±4 2
)︃𝑛
.
We used a program written in Maple to see that𝑛1= 1in all cases. Here,𝑃𝑛(𝑋)is one of the two polynomials giving𝑥𝑛 in terms of𝑥1for the equation𝑥2−𝑑𝑦2=±4.
From equation (4.8), for𝑖= 2we get
⃒⃒
⃒⃒𝑛2
log𝛿
log 10− log𝑏2
log 10−𝑚2
⃒⃒
⃒⃒< 12
(log 10)10𝑚2, (4.18) where 𝛿=𝑥1+𝑦1
√𝑑=𝑥1+√︀
𝑥21±4,𝑥1=𝑎1(10𝑚1−1)/9, and𝑏2 =𝑎2/9 with 𝑎1̸=𝑎2. To apply Lemma 3.1 to inequality (4.18), we put
𝜅= log𝛿
log 10, 𝜇= log𝑏2
log 10, 𝐴= 12
log 10, 𝐵= 10, and 𝑀= 1.5·1029. The program was developed in PARI/GP running with 200 digits, for1 ≤𝑚1 ≤ 214. For the computations, if the first convergent such that 𝑞 > 6𝑀 does not satisfy the condition 𝜂 >0, then we use the next convergent until we find the one that satisfies the conditions. In a few minutes, all the computations were done. In all cases, after the first run we obtained𝑚2≤35. We set 𝑀= 35and the second run of the reduction method yields𝑚2≤8. In conclusion, we have
𝑛1= 1, 1≤𝑚1≤8, 1≤𝑚2≤8, 1≤𝑛2≤48.
Now a verification by hand yields the final result.
Acknowledgements. F. L. was supported in part by grant CPRR160325161141 and an A-rated scientist award both from the NRF of South Africa and by grant no. 17-02804S of the Czech Granting Agency. This paper was finalized during a visit of A.T. at the School of Mathematics of Wits University in August 2017.
This author thanks this institution for its hospitality and the CoEMaSS at Wits for support.
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